Question: Definition: If f(x) is a function, then we say that a value u is a fixed point of f(x) if ar only if f(u)=u .

Definition: If

f(x)

is a function, then we say that a value

u

is a fixed point of

f(x)

if ar only if

f(u)=u

.\ Suppose

F(x)

is a given continuous function and

a!=0

is a given real number.\ a. Show that

u

is a zero of

F(x)

if and only if

u

is a fixed point of

f(x)

f(x)=x+aF(x)

\ b. Suppose

F^(')(x)

is continuous,

u

is a zero of

F

, and

F^(')(u)!=0

. Define

f(x)=x+aF(x)

. Prove there are values of

a!=0

and

\\\\epsi >0

so that if

u_(0)in

(u-\\\\epsi ,u+\\\\epsi )

and

u_(n+1)=f(u_(n))

for

n=0,1,2,dots

then

u_(n)->u

as

n->\\\\infty

. Hint:

|u_(n+1)-u|=|f(u_(n))-f(u)|

. Use the definition of

f(x)

and the mean value theorem.

 Definition: If f(x) is a function, then we say that a

Definition: If f(x) is a function, then we say that a value u is a fixed point of f(x) if a only if f(u)=u. Suppose F(x) is a given continuous function and a=0 is a given real number. a. Show that u is a zero of F(x) if and only if u is a fixed point of f(x) f(x)=x+aF(x) b. Suppose F(x) is continuous, u is a zero of F, and F(u)=0. Define f(x)=x+aF(x). Prove there are values of a=0 and >0 so that if u0 (u,u+) and un+1=f(un) for n=0,1,2, then unu as n. Hint: un+1u=f(un)f(u). Use the definition of f(x) and the mean value theorem

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